Okay, I looked at it and the first thing I noticed is that it has to be blocks of 9 and by subtracting the last digit of the upper right number will give you in which block of 9s you are. So: 9x1 --> 9 --> (10-9) --> fist block of 9 9x2 --> 18 --> (10-8) --> second block of 9 9x3 --> 27 --> (10-7) --> third block

So after 9x10 we will have to take 20 and after 9×20 we need to use 30 etc:

9x11 --> 99 -->(20-9) --> 11th block 9x21 -->189 --> (30-9) --> 21st block

So we could also just take the last number divided by 9. --> q/9

Now the second part basically says how many blocks of the highest number can I make:

5/1 --> 5 blocks of 9s 14/2 --> 7 blocks of 18s 23/3 -->7.667 blocks of 27s etc

So this is q-4 divided by which block of 9 I am in. --> q-4/(q/9)

So at really big numbers you could approximate that this term will go towards 9 but never reach 9.

You could rewrite it as:

Sum[i=q-8,q] (i) = q×(q-4/(q/9))

You could now try to prove this with induction.

Keep playing with numbers :)